At equilibrium in a 10.0 L vessel,there are 7.6 × 10^-2 moles of SO₂,8.6 × 10^-2 moles of O₂,and 8.2 × 10^-2 moles of SO₃.What is the equilibrium constant (K_c)for the reaction:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the reaction is as follows: K_c = [SO₃]2 / [SO₂]2[O₂] Substituting the given equilibrium concentrations: K_c = [(8.2 × 10^-2) / 10.0]2 / [(7.6 × 10^-2) / 10.0]2[(8.6 × 10^-2) / 10.0] Simplifying this expression gives us: K_c = 135 Therefore, the correct answer is (C).

At Equilibrium in a 10